Show $\lim(s_n) = +\infty$ if and only if $\lim(-s_n) = -\infty$. Show $\lim(s_n) = +\infty$ if and only if $\lim(-s_n) = -\infty$.
I'm not really sure what to do here. I think I use limit laws to prove the problem, but how do I do that in this case?
 A: $\lim s_n=+\infty$ $\iff$ 
For all $M$ there is $N$ such that if $n>N$ then $s_n>M$. $\iff$
For all $M$ there is $N$ such that if $n>N$  then $-s_n<-M$. $\iff$
For all $M$ there is $N$ such that if $n>N$  then $-s_n<M$. $\iff$
$\lim -s_n=-\infty$.
A: The definition of $\lim_n s_n=+\infty$ is that for all $\epsilon>0$, there exists an $N$ such that $s_n>\epsilon$ for all $n\geq N$. The formulation for $-\infty$ is similar. 
So assume $\lim_n s_n=+\infty$ and work out that $\lim_n (-s_n)=-\infty$. Then do it the other way around.
A: Here's the point of the problem:
You need to prove if $\{ s_{n} \}$ is a sequence, then the following is true:
$$\lim \limits_{n \to \infty} s_{n} = \infty \iff \lim \limits_{n \to \infty} -s_{n} = -\infty$$
and that means you have to prove two statements:
1)  if $\lim \limits_{n \to \infty} s_{n} = \infty$, then $\lim \limits_{n \to \infty} -s_{n} = -\infty$ 
and 
2) $\lim \limits_{n \to \infty} -s_{n} = -\infty$, then $\lim \limits_{n \to \infty} s_{n} = \infty$.
Now, as we noted, what it means for $\lim \limits_{n \to \infty} s_{n} = \infty$ is $\forall \epsilon > 0$, $\exists N$ such that $n \geq N$ implies $s_{n} > \epsilon$.
And, what it means for $\lim \limits_{n \to \infty} - s_{n} = -\infty$ is $\forall \epsilon > 0$, $\exists N$ such that $n \geq N$ implies $-s_{n} < - \epsilon$.
Ok, so to prove 1), let's suppose $\lim \limits_{n \to \infty} s_{n} = \infty$.  Then that means $\forall \epsilon > 0$, $\exists N$ such that $n \geq N$ implies $s_{n} > \epsilon$.  But if this is true, then multiplying $s_{n} > \epsilon$ on both sides by $-1$ gives $-s_{n} < - \epsilon$, and this is true for all $n \geq N$.  But that's exactly what it means for $\lim \limits_{n \to \infty} -s_{n} = -\infty$.
You can prove 2) in the exact same way, just reversed.
